By documentation you see that random.shuffle shuffles the list itself and does not return no value:
(..) Shuffle the sequence x in place. (...)
Starting from a concrete example of a list of notes:
l = [15,2,20,12]
If you do:
shuffle(l)
You're shuffling the list. And depending on how it was shuffled it could look like this:
>>> l
[2, 15, 12, 20]
The function, however, does not return any value or any new shuffled list, it only changes the list received by parameter. So it will not make sense to do:
.format(shuffle (l) )
The solution to your code is exactly as you mentioned:
from random import shuffle
n1= str (input ('Digite o nome do primeiro aluno: '))
n2= str (input ('Digite o nome do segundo alino: '))
n3= str (input ('Digite o nome do terceiro aluno '))
n4= str (input ('Digite o nome do quarto aluno: '))
l= [n1,n2,n3,n4]
shuffle (l) # só shuffle
print ('A ordem dos trabalhos será: {}' .format(l)) # só mostrar
You can even see the implementation of the shuffle function by referring to your source code :
def shuffle(self, x, random=None):
"""Shuffle list x in place, and return None.
Optional argument random is a 0-argument function returning a
random float in [0.0, 1.0); if it is the default None, the
standard random.random will be used.
"""
if random is None:
randbelow = self._randbelow
for i in reversed(range(1, len(x))):
# pick an element in x[:i+1] with which to exchange x[i]
j = randbelow(i+1)
x[i], x[j] = x[j], x[i]
else:
_int = int
for i in reversed(range(1, len(x))):
# pick an element in x[:i+1] with which to exchange x[i]
j = _int(random() * (i+1))
x[i], x[j] = x[j], x[i]
Notice how nowhere else in this function is a return made.
Notice the comment made at the beginning of the function:
"" Shuffle list x in place, and return None.
